Problem 83 - Intermediate Algebra

Question

Solve for the specified variable. $$ S=\frac{n}{2}[2 a+(n-1) d] \quad \text { for } d $$

Step-by-Step Solution

Answer

The solution for \( d \) is \( d = \frac{2(S - na)}{n(n-1)} \).

Step 1: Isolate the brackets

Start by expanding the brackets in the equation. The original equation is \( S = \frac{n}{2}[2a + (n-1)d] \). Expanding the brackets gives:\[ S = \frac{n}{2} \cdot 2a + \frac{n}{2}\cdot (n-1) d \]

Step 2: Simplify the equation

Simplify the equation by distributing the terms:\[ S = na + \frac{n(n-1)}{2}d \]

Step 3: Isolate terms involving d

To solve for \( d \), we need to isolate terms involving \( d \). Subtract \( na \) from both sides to get:\[ S - na = \frac{n(n-1)}{2}d \]

Step 4: Solve for d

Now, solve for \( d \) by dividing both sides by \( \frac{n(n-1)}{2} \):\[ d = \frac{2(S - na)}{n(n-1)} \]

Key Concepts

Algebraic Manipulation

Algebraic manipulation is a crucial aspect of solving mathematical equations. It involves rearranging equations and expressions to isolate specific terms or variables. This process requires applying basic arithmetic operations such as addition, subtraction, multiplication, and division to both sides of an equation.

• Expanding Brackets: One common application of algebraic manipulation is expanding brackets. In the given equation, the brackets \[2a + (n-1)d\] were expanded to \[ 2a + (n-1)d.\]
• Distributive Law: While expanding the brackets, the distributive law is used to multiply each term inside the brackets by the term outside of it. This law states that \((x + y)z = xz + yz\).

Mastering algebraic manipulation allows you to transform complex equations into simpler forms that are easier to solve.

Variables

Variables are symbols, often letters, that represent numbers in equations and expressions. They serve several roles in algebra:

• Representing Unknowns: Variables like \( d \) in the exercise are placeholders for values that need to be found.
• Maintaining Generality: Using variables allows equations to remain general and applicable in various scenarios, not limited to specific numbers.
• Flexibility: They enable the formulation of relationships between quantities, such as \( n \) and \( a \) in the provided equation.

When working with equations, understanding the role of each variable is essential to accurately solving the equation and interpreting its meaning.

Equation Solving

Equation solving involves finding the value of a variable that makes an equation true. It is a step-by-step process where each step is guided by mathematical operations and properties:

• Isolating the Variable: The primary goal is to isolate the target variable on one side of the equation, as seen in the original exercise where \( d \) is solved for.
• Balancing the Equation: This means performing the same operation on both sides of the equation to maintain equality, such as subtracting \( na \) from both sides.
• Simplifying the Equation: Simplifying both sides of the equation can make the isolation of the variable straightforward, turning complex expressions into simpler forms.

Successful equation solving requires careful application of algebraic rules at each step.

Intermediate Algebra

Intermediate algebra bridges the gap between basic algebra and more advanced mathematical topics. It involves solving more complex equations and manipulating expressions that have multiple terms and variables, as demonstrated in this exercise.

• Understanding Complex Equations: Problems in intermediate algebra often include equations, like the given one, which need careful manipulation and multiple steps to solve.
• Using Distributive and Associative Properties: These properties help in expanding and simplifying expressions effectively, essential for handling more advanced equations.
• Applying Critical Thinking: Solving intermediate algebra problems often requires strategic thinking to choose the right operations to simplify and solve equations.

Understanding intermediate algebra concepts can significantly enhance problem-solving skills, making it easier to tackle more advanced mathematical challenges.